Question

The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by y= (4t- 2t^2)m and
x = (3t) metre, where t is in second and point of
projection is taken as origin. The angle of projection of projectile with vertical is
(1) 30°
(2) 37°
(3) 45°
(4) 60°

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Positions are given by:-

$\large{\bold{X = 3t}}$ and,

$\large{\bold{y = 4t - 2t^2}}$

Let the angle of projection be $\large{ \theta}$ with vertical.

$\huge{\bold{\underline{Explanation:-}}}$

As the Displacement of the body are given

We need to find velocities in respective axis ( x and y axis)

Taking y - coordinate displacement and differentiating it.

$\large{\bold{y = 4t - 2t^2}}$

Differentiating w.r.t time

$\large{v_y = \frac{dy}{dt}}$

Substituting the values.

$\large{ \implies v_y = \frac{d(4t - 2t^2)}{dt}}$

It comes as,

$\large{\boxed{v_y = 4 -4t}}$

But we know $\large{v_y = v \sin \theta}$

So, equation becomes,

$\large{ \implies v \sin \theta = 4 - 4t}$

At time of projection time "t" = 0 seconds.

Substituting the time value in above equation.

$\large{ \implies v \sin \theta = 4 - 4 \times 0}$

$\large{v \sin \theta = 4 \: m/s \: -------(1)}$

Taking y - coordinate displacement and differentiating it.

$\large{\bold{x = 3t}}$

Differentiating w.r.t time

$\large{v_x = \frac{dx}{dt}}$

Substituting the values.

$\large{ \implies v_x = \frac{d(3t)}{dt}}$

Simplifying,

$\large{\boxed{v_x = 3}}$

But we know $\large{v_x = v \cos \theta}$

So, equation becomes,

$\large{ \implies v \cos \theta = 3}$

As here we can see velocity of x - coordinate is not dependent of time I.e it remains constant throughout the motion of the body.

$\large{v \cos \theta = 3 \: m/s \: -------(2)}$

Now,

Divide equation (1) and (2) .

$\large{ \implies \frac{v \sin \theta}{v \cos \theta} = \frac{4}{3}}$

$\large{ \implies \frac{ \cancel{v} \sin \theta}{ \cancel{v} \cos \theta} = \frac{4}{3}}$

$\large{ \implies \frac{ \sin \theta}{ \cos \theta} = \frac{4}{3}}$

As we know sinθ/cosθ = tanθ.

Substituting the values,

$\large{ \implies \tan \theta = \frac{4}{3}}$

$\large{ \implies \tan \theta = \tan 37 \degree}$

$\huge{\boxed{\boxed{ \theta = 37 \degree}}}$

So, the angle of projection with vertical is 37°(Option - 2).